Question

A stone thrown upward from the top of a tower 85m high reaches the ground in 5 second 1:-calculate the greatest height above the ground? 2:-the velocity with which it reaches the ground? 3:-the time taken to reach the aximum height?(g=10ms sq)

Answer 1

Solution:

A stone is thrown in the upward direction from the top of the tower and it reached the ground in 5s

Let the initial velocity be ‘u’

            The displacement is ‘-85’ and acceleration is ‘-10 $m/s^{2}$’ as because the stone is coming to the ground (downward direction) so the negative sign comes.

Using the formula,

$s=u t+\frac{1}{2} a t^{2}$

$s-85=5 u-\frac{1}{2} \times 10 \times 52$

$u = 8 m/s$

 At the maximum height above the level of ground, the stone comes to momentary rest. That means its velocity becomes zero.

Using the formula,

$v^{2}=u^{2}+2 a s$

$p=8^{2}-(2 \times 10 \times s)$

$-64=-20 s$

s=3.2m

Thus, the maximum height from the ground is = 3.2 + 85 = 88.2 m.

Velocity with which the stone reaches the ground is,  

$v=u+a t$

$v=8-10 \times 5$

v = -42 m/s [downward]

-42 m/s is the value of velocity with which the stone reaches the ground.

The time taken for reaching the maximum height,$t_{1}$

$t_{1}=\frac{u_{1}}{a}$

=$\frac{8}{10}$

=0.8 s

The time of 0.8 second is required for reaching the maximum height.

Answer 2

Explanation:

A stone is thrown in the upward direction from the top of the tower and it reached the ground in 5s

Let the initial velocity be ‘u’

The displacement is ‘-85’ and acceleration is ‘-10 m/s^{2}m/s

2

’ as because the stone is coming to the ground (downward direction) so the negative sign comes.

Using the formula,

s=u t+\frac{1}{2} a t^{2}s=ut+

2

1

at

2

s-85=5 u-\frac{1}{2} \times 10 \times 52s−85=5u−

2

1

×10×52

u = 8 m/su=8m/s

At the maximum height above the level of ground, the stone comes to momentary rest. That means its velocity becomes zero.

Using the formula,

v^{2}=u^{2}+2 a sv

2

=u

2

+2as

p=8^{2}-(2 \times 10 \times s)p=8

2

−(2×10×s)

-64=-20 s−64=−20s

s=3.2m

Thus, the maximum height from the ground is = 3.2 + 85 = 88.2 m.

Velocity with which the stone reaches the ground is,

v=u+a tv=u+at

v=8-10 \times 5v=8−10×5

v = -42 m/s [downward]

-42 m/s is the value of velocity with which the stone reaches the ground.

The time taken for reaching the maximum height,t_{1}t

1

t_{1}=\frac{u_{1}}{a}t

1

=

a

u

1

=\frac{8}{10}

10/8

=0.8 s

The time of 0.8 second is required for reaching the maximum height